Glasgow Math. J.53(2011) 523–534.Glasgow Mathematical Journal Trust 2011.
doi:10.1017/S0017089511000103.
THE CONVEX INTERSECTION BODY OF A CONVEX BODY
MATHIEU MEYER
Universit´e Paris-Est - Marne-la-Vall´ee, Laboratoire d’Analyse et de Math´ematiques Appliqu´ees (UMR 8050), Cit´e Descartes, 5 Bd Descartes, Champs-sur-Marne, 77454 Marne-la-Vall´ee cedex 2, France
e-mail: mathieu.meyer@univ-mlv.fr
and SHLOMO REISNER†
Department of Mathematics, University of Haifa, Haifa 31905, Israel e-mail: reisner@math.haifa.ac.il
(Received 2 April 2010; revised 29 September 2010; accepted 30 November 2010;
first published online 10 March 2011)
Abstract. LetLbe a convex body in⺢nandzan interior point ofL. We associate withLandza new, convex and centrally symmetric, bodyCI(L,z). This generalizes the classicalintersection body I(L,z) (whose radial function atu∈Sn−1is the volume of the hyperplane section ofLthroughz, orthogonal tou).CI(L,z) coincides with I(L,z) if and only if L is centrally symmetric aboutz. We study the properties of CI(L,z).
2010Mathematics Subject Classification.52A20.
1. Introduction. LetLbe a convex body in⺢ncontaining 0 in its interior. The intersection bodyI(L) of L, defined by its radial functionρI(L) on the sphereSn−1, which is
ρI(L)(u)=vol(L∩u⊥) and the cross-section bodyC(L) ofL, defined by
ρC(L)(u)=max
t vol((L∩(tu+u⊥))
are not, in general, convex bodies, although they are identical and, moreover, convex in the case whenLis centrally symmetric. This follows from Brunn–Minkowski theorem (see [21, p. 309]) and from Busemann’s theorem (see [3]). We introduce here a new convex body associated withL, generalizing both the intersection body and the cross- section body. More precisely, we define theconvex intersection body CI(L) ofLby its radial function
ρCI(L)(u)= min
z∈Pu(L∗g(L))vol
Pu(L∗g(L))∗z
, (1)
†Both authors were supported in part by the France-Israel Research Network Program in Mathematics contract #3-4301.
whereg(L) denotes here the centroid ofL, and ifEis a linear subspace of⺢nandM is a convex body inE, we define forz∈E,
M∗z= {y∈E;y−z,x−z ≤1 for everyx∈M}.
Thus (1) means: first apply duality with respect to the pointg(L), then project ontou⊥, finally apply duality with respect tozand minimize the (n−1)-dimensional volume overz.
In Section 2, we shall attach to any convex bodyKin⺢na bodyJ(K) constructed with the help of its projections, and prove that it is always convex. In Section 3, we prove thatCI(L) is convex, and we study the inclusions
CI(L)⊂I(L)⊂C(L),
whengL=0. Finally, in Section 4, a few open problems are listed, with proposed ideas concerning some of them.
NOTATIONS. Forx,y∈⺢n, we denote byx,ythe canonical scalar product in⺢n,
|x|denotes the Euclidean norm defined by it. Ifuandvare two non-zero vectors in⺢n that are not orthogonal to one another, we definew,u⊥ :⺢n→u⊥to be the projection parallel towontou⊥= {x∈⺢n;x,u =0}, we denote
Pu=u,u⊥.
IfLis a subset of⺢n, let [L] be the affine subspace of⺢nthat it spans. IfBis a convex subset of⺢n, we denote its k-dimensional volume by vol(B) (wherek=dim[B]). By conv (A), we denote the closed convex hull ofA.
ForLa convex set in⺢nandz∈[L], we denote the polar body ofLwith respect tozby
L∗z= {y∈[L];y−z,x−z ≤1 for everyx∈L}.
It is well known that the functionz→vol(L∗z) is strictly convex on the relative interior ofLand tends to+∞aszapproaches the realtive boundary ofLin [L] (see [17,20]). It follows that it reaches its minimum at a unique points(L)∈int(L). This point is called theSantal´o point of L. We shall denote
L∗s:=L∗s(L).
Moreover,s(L) is also characterized as the unique pointz∈int(L), which is the centroid ofL∗z(see [20] and also [21, p. 419]). Let us denote byg(M) the centroid of a convex bodyMin [M], and set
M∗g=M∗g(M). One has
L∗s=Mif and only ifM∗g=L.
Observe that, in general, s(L)=g(L) (see the recent [18], where a lower bound to how far apart these two points can be is given). Finally if 0∈int(M), we shall write M∗ =M∗0, so thatM∗x=x+(M−x)∗for everyx∈int(M).
We adopt the following notation: ifKis a star body with respect to 0, we denote
||x||K =inf{λ >0;x∈λK}
to be thegaugeofK. Then foru∈Sn−1, ρK(u)= 1
||u||K
is theradial functionofK.
2. A convexity theorem.
THEOREM1. If K is a convex body in ⺢n, let NK: ⺢n→⺢+ be defined by the formula
NK(u)= 1
vol((PuK)∗s) = 1
minz∈u⊥vol((PuK)∗z)for u∈Sn−1,
and extended to all⺢nby NK(ru)=rNK(u)for r≥0and u∈Sn−1. Then NKis a norm on⺢n.
Before proving Theorem 1, we need some preliminary results.
DEFINITION. Let v∈Sn−1, B be a bounded subset of ⺢n and V:B→⺢ be a bounded map. Theshadow system(Lt),t∈[a,b] of convex bodies in⺢n, withdirection the vectorv, withbasisthe setBand withspeed the functionV, is the family of convex bodies
Lt=conv{b+tV(b)v;b∈B}, fort∈[a,b].
For the sake of completeness, we prove the following result, which appears in [23] and is used, for example, in [4,5].
PROPOSITION 2.Let K be a convex body in ⺢n. Then, for u, v∈Sn−1, such that u, v =0, the family Lt=u+tv,u⊥K, t∈⺢, is a shadow system of convex bodies in u⊥, in the directionv.
Proof.To simplify notation, one may suppose that u=en andv=en−1, where e1, . . . ,enis an orthonormal basis of⺢n (we shall write⺢jfor [e1, . . . ,ej]). Then, for allt∈⺢,
u+tv,u⊥K = {X+zen−1;X∈⺢n−2,X+zen−1+r(en+ten−1)∈Kfor somer∈⺢}
= {X+zen−1;X∈⺢n−2,X+(z+rt)en−1+ren∈Kfor somer∈⺢}
= {X+(x−rt)en−1; X∈⺢n−2,r∈⺢such thatX+xen−1+ren∈K}
= {U−rten−1; (U,r)∈PuK×⺢such thatU+ren∈K}.
ForU∈PuK, define
I(U)= {r∈⺢;U+ren∈K}.
ThenI(U)=[a(U),b(U)] is a closed interval of⺢. Define alsox(U)∈⺢such that U−x(U)en−1,en−1 =0,
and let
D1= {U∈PuK;x(U)∈⺡}andD2= {U∈PuK;x(U)∈⺢\⺡}.
DefineV :PuK→⺢by
v(U)= −b(U) ifU∈D1andv(U)= −a(U) ifU∈D2.
By the continuity of the two concave functions−a,b:Pu(K)→⺢, it is easy to see that for everyt∈⺢
u+tv,u⊥K=conv{U+tV(U)en;U∈PuK}.
REMARK. The converse assertion of Proposition 2 is true : every shadow system Lt in ⺢n can be represented as Lt=u+tv,u⊥(K) with an appropriate convex body K⊂⺢n+1andu, v∈Sn. This was shown, for example, in [4].The following result was proved in [16].
THEOREM3.Let t∈[a,b]→Lt be a shadow system in⺢n, then the functionφ:
⺢n→⺢, defined by the formula φ(t)= 1
vol((Lt)∗s) = 1
minzvol((Lt)∗z), is convex.
LEMMA4.Let N:⺢n→⺢+satisfy
r
N(x)>0for x=0,r
N(αx)= |α|N(x)for everyα∈⺢and x∈⺢n,r
for all u, v∈Sn−1such thatu, v =0, t→N(u+tv)is convex on⺢. Then N is a norm on⺢n.Proof. Let us show that N(x+y)≤N(x)+N(y) for every x,y∈⺢n\ {0}. Let α= ||x|x +|y|y|andβ= ||x|x −|y|y|,u= 1α(|x|x +|y|y ) andv= β1(|x|x −|y|y). We may suppose thatα=0 andβ=0. Thenu, v∈Sn−1,u, v =0 and
x= |x|
2 (αu+βv), y= |y|
2 (αu−βv). We get
N(x+y)= α(|x| + |y|)
2 N
u+β(|x| − |y|) α(|x| + |y|)v
= α(|x| + |y|)
2 N(u+(λt+(1−λ)s)v),
where λ= |x|+||x|y|, t= βα and s= −βα. Under the assumption of the lemma, we get
N(x+y)≤ α(|x| + |y|)
2 (λN(u+tv)+(1−λ)N(u+sv))
= α 2
|x|N
u+β αv
+ |y|N
u−β
αv
=N(x)+N(y).
Proof of Theorem 1.In view of Lemma 4, we need to prove that t→gu,v(t)= N(u+tv) is convex, wheneveru, v∈Sn−1 satisfyu, v =0. It is easy to see that for anyt∈⺢,Pu+tvKis an affine image ofu+tv,u⊥Kand satisfiesvol(Pu+tvK)= 1
√1+t2vol(u+tv,u⊥K).
Hence
min
z∈{u+tv}⊥vol((Pu+tvK)∗z)=
1+t2min
z∈u⊥vol((u+tv,u⊥K)∗z)) It follows that
N(u+tv)= |u+tv|
minz∈{u+tv}⊥vol((Pu+tvK)∗z)) = 1
minz∈u⊥vol((u+tv,u⊥K)∗z) . Now by Proposition 2,t→u+tv,u⊥Kis a shadow system on⺢and thus by Theorem 3, gu,vis convex on⺢.
REMARKS. (1) IfKis centrally symmetric (and centred at 0), then all its projections PuKare centrally symmetric (and centred at 0) so that
minz∈u⊥vol(PuK)∗z)=vol(PuK)∗0)=vol(K∗0∩u⊥).
Under this hypothesis, we proved that u→ vol(K1∗0∩u⊥) is the restriction to Sn−1 of a norm on ⺢n. This is Busemann [3] theorem on the sections of convex centrally symmetric bodies, applied toK∗.
(2) For every convex bodyKin⺢n, Theorem 1 defines a centrally symmetric convex bodyJ(K) in⺢nby
J(K)= {x∈⺢n;NK(x)≤1}.
Notice that for everyx0∈⺢n, J(K+x0)=J(K) and that forA:⺢n→⺢n a linear isomorphism, we have
J((AK))= |det(A)|(A∗)−1(J(K)). (3) Ifn=2 and ifRis the rotation by angleπ/2 in⺢2, then
vol(PuK)=hK(Ru)+hK(−Ru)=hK(Ru)+h−K(Ru),
so that
J(K)= 14R(K−K).
3. The convex intersection bodiesIC(L,z)of a convex bodyL. LetLbe a convex body in⺢n. For a point z∈int(L), theintersection body I(L,z)of L with respect to zis the centrally symmetric star body in⺢n whose radial functionρI(L,z) is given for u∈Sn−1by
ρI(L,z)(u)=vol({x∈L;x−z,u =0})=vol(L∩(z+u⊥)). The bodyC(L) is the star body in⺢ndefined by its radial function
ρC(L)(u)=max
x∈L vol(L∩(x+u⊥)).
Of course, one hasI(L,z)⊂C(L) for everyz∈int(L). It was proved in [13] that these bodies coincide if and only ifLis centrally symmetric aboutz(the ‘if’ part follows easily from Brunn–Minkowski theorem). We define now theconvex intersection body of L with respect to z∈int(L), which we denote byCI(L,z), by
CI(L,z)=J(L∗z).
Whenz=g(L) is the centroid ofL, we shall denoteCI(L)=CI(L,g(L)). The radial function ofCI(L,z) is thus given foru∈Sn−1by
ρCI(L,z)(u)=min
x∈u⊥vol((Pu(L∗z))∗x)=vol((Pu(L∗z))∗s). In view of Theorem 1, one has
THEOREM5.Let L be a convex body. Then for every z∈int(L), CI(L,z)is a centrally symmetric convex body such that CI(L,z)⊂I(L,z).
REMARKS. (1) It is easy to see that one has for every one-to-one affine map A:⺢n→⺢n,I(AL,Az)= |det(A)|A∗−1
I(L,z)
, as well as CI(AL,Az)= |det(A)|A∗−1
CI(L) .
(2) In the casen=2, foru∈S1, denoting byRthe rotation around 0 of angleπ/2, one has
||u||C(L)= ||Ru||K−K
≤ ||u||I(L,z)= 1
||Ru||K−z+ 1
|| −Ru||K−z) −1
≤ ||u||CI(L,z)=4(||Ru||K−z+ || −Ru||K−z).
(3) The inclusionCI(L,z)⊂I(L,z) is exact in the sense that their boundaries touch;
there always existsu∈Sn−1such that
vol(L∩(z+u⊥))=vol((Pu(L∗z))∗s).
As a matter of fact, this equality means that the centroid ofL∩(z+u⊥) inz+u⊥is atz. To see that suchuexists, defineφ:Sn−1→⺢by
φ(v)=vol({x∈L;x−z;v ≥0}).
Sinceφis continuous, it reaches its maximum at some pointu∈Sn−1. Then, by [15],z is the centroid ofL∩(z+u⊥).See also [10].
(4) It was proved by Gr ¨unbaum ([10, Section 6.2]) that for every convex body L∈⺢n, there exists somez0 ∈int(L) such that (n+1) different hyperplanes through z0, with normalsu1, . . . ,un+1, satisfy thatz0is the centroid ofL∩(z+u⊥i ). For thisz0, the boundaries ofCI(L,z0) and ofI(L,z0) have at least 2(n+1) contact points.
(5) We have seen above that, in the case whenLis centrally symmetric aboutz, CI(L,z)=I(L,z) and Theorem 5 is nothing else but the classical Busemann’s theorem [1]. Conversely, the following result holds.
PROPOSITION6.One has CI(L,z)=I(L,z)if and only if L is centrally symmetric about z.
It is a consequence of the following lemma.
LEMMA7.Let L be a convex body and z∈L. Suppose that z is the centroid of every hyperplane section of L through itself. Then L is centrally symmetric about z.
Proof.Fix somez0∈int(L),z0=z, and defineF:⺢n→⺢by F(y)=vol({x∈L−z0;x,y ≥1}). By [15],FisC1on{F>0} =⺢n\ {0}and one has fory=0
∇F(y)= ∇F(y),yg({w∈L−z0;w,y =1})
= ∇F(y),y
g({x∈L;x−z0,y =1})−z0
. (2)
LetHbe the affine hyperplane in⺢ndefined by
H = {y∈⺢n;z−z0,y =1}.
If y∈H, the hyperplane {x∈⺢n;x−z0,y =1}passes through z, so that by the hypothesis, one has
g({w∈L;x−z0,y =1})=z, thus, by (2) we get
∇F(y)= ∇F(y),y(z−z0). Now ify,y∈H, one has
y−y,z−z0 =0 and for everyt∈⺢, (1−t)y+ty∈H so that
F(y)−F(y)= 1
0
y−y,∇F((1−t)y+ty))dt=0.
Thus,Fis equal to some constantconH. Define a functionG:Sn−1 →⺢by G(u)=vol{x∈L;x−z,u ≥0}.
and let
U= {u∈Sn−1;u,z−z0>0}.
Thenu→y(u) :=u,zu−z0 is a one-to-one mapping fromUontoH, and it is easy to check that
G(u)=F y(u)
for everyu∈U.
It follows thatG=conU, and sinceG(u)+G(−u)=vol(L) for everyu∈Sn−1,G= vol(L)−con−U. Now,Sn−1∩(z−z0)⊥ is contained in the closures of bothUand of−U inSn−1. SinceGis continuous onSn−1,G=c=1−c= vol(L)
2 on Sn−1. The fact thatL is centrally symmetric aboutznow follows by a classical result (see [8]
or [6]).
4. Additional comments and some open problems. We know that althoughC(L) andI(L,z) are not in general convex bodies (by [14],C(L) is convex forn≤3 and by [2], if nis the simplex in⺢n,C( n) is not convex ifn≥4). HoweverC(L) andI(L,g(L)), whereg(L) is the centroid ofL, arealmost convex, and evenalmost ellipsoids, in the sense that there exist some constantsc>d >0, independent onnandL, such that for everyu∈Sn−1, one has
d
vol(L)32 L−g(L)
x,u2dx 12
≤ 1
maxtvol
L∩(tu+u⊥) =ρC(L)(u)
≤ 1
vol(L∩u⊥) =ρI(L,g(L))(u)
≤ c
vol(L)32 L−g(L)
x,u2dx 12
.
In the centrally symmetric case, this was proved by Hensley, and Ball [1] (for sharp constants, see also [19]), and in the general case by Sch ¨utt [22] and Fradelizi [7] (the latter with sharp constants). We have seen thatρI(L,g(L))≤ρCI(L,g(L)). A natural question to ask now is
OPEN PROBLEM1. Does there exist a universal constantC>0, independent on the convex bodyLin⺢n and onn≥1, such thatρCI(L,g(L))≤CρI(L,g(L))? Of course, in view of the preceding inequalities, an affirmative answer to this question would say that the radial functions ofC(L),CI(L) andI(L,g(L)) are all equivalent (with absolute constants).
Observe that an equivalent way of formulating this problem is the following. LetK be a convex body in⺢nsuch that its Santal ´o point is at 0. Does there exist an absolute constantC>0, independent onnandKsuch that
vol
(PuK)∗Puz
≥Cvol
(PuK)∗0
for everyz∈int(K) ?
Equivalently, given a convexM⊂u⊥, with Santal ´o points(M), and a convex bodyK in⺢n, with Santal ´o points(K), such thatPuK=M, does
vol(M∗s(M))≥Cvol(M∗Pus(K)) for some universal constantC>0 ?
If one could prove that in this situation, for some universal constantc>0, the following is true:
Pus(K)−s(M)∈ c n
M−s(M) ,
then an affirmative answer could be given, using the following lemma.
LEMMA8.Let V be a convex body in⺢nand x,y∈int(V). Then (1− ||x−y||V−y)nvol(V∗x)≤vol(V∗y)≤ vol(V∗x)
1− ||y−x||V−x
n
Proof. One has
vol(V∗y)=vol(V∗y+y)=vol
(V−y)∗
=vol(
(V−x−(y−x))∗
= (V−x)∗
1
(1− y−x,z)n+1dz
because by a formula given in [18], ifLis a convex body with 0 in its interior, one has for everywin the interior ofL,
vol(L∗w)=
L∗
1
(1− w,z)n+1dz. Sincey−x,z ≤ ||z||(V−x)∗||y−x||V−x, we get
vol(V∗y)≤
(V−x)∗
1
(1− ||z||(V−x)∗||y−x||V−x)n+1dz. (3) Now, ifLis a convex body with 0 in its interior, and 0<c<1, then
L
1
(1−c||z||L)n+1dz=nvn Sn−1
||θ||L1
0
rn−1
(1−cr||θ||L)n+1dr
dθ
=nvn Sn−1
1 n
r 1−cr||θ||L
n||θ||L1
0
dθ
=vn Sn−1
1
(1−c)n||θ||nLdθ= vol(L) (1−c)n. From which, together with (3), we get
vol(V∗y)≤ vol(V∗x) (1− ||y−x||V−x)n.
Applying the same formula while interchanging the roles ofxandy, one has vol(V∗x)≤ vol(V∗y)
(1− ||x−y||V−y)n.
It is well known (see, e.g. [19]) that there is an affine transformationA:⺢n→⺢nsuch thatAL=Misisotropic, that is, it satisfies vol(M)=1 and for everyu∈Sn−1,
1
vol(M) M−g(M)x,u2dx 12
=LM,
whereLMisthe isotropic constantofM. In that context, Problem 1 is equivalent to OPENPROBLEM2. LetMbe an isotropic convex body. IsCI(M) equivalent to the Euclidean ball (with absolute constant independent onM⊂⺢nand onn) ?
Of course, Problems 1 and 2 are non-trivial only ifL or M are not centrally symmetric. The particular case of the simplex is still open.
OPENPROBLEM3. Let nbe a simplex in⺢n. Is there a constantcindependent on nsuch that for everyu∈Sn−1
vol( n∩u⊥)≤cvol((( n∩u⊥)∗0)∗s)=cvol((Pu( ∗ng))∗s)?
Observe that when n is a regular simplex inscribed in the Euclidean ball, since ( n)∗0= −n n, one has
( n∩u⊥)∗0=Pu(( n)∗0)=Pu(−n n) and thus
vol((( n∩u⊥)∗0)∗s)= 1
nn−1vol((Pu n)∗s).
About Problem 3, one may remark that by affine invariance, we may suppose without loss of generality that n is the regular simplex with vertices e1, . . . ,en+1,|ei| =1, centred at 0=e1+ · · · +en+1. For 1≤i=j≤n+1, one has thenei,ej = −1n.
FACT.Let A⊂ {1, . . . ,n+1}be such that1≤k:=card(A)≤n and define uA=
i∈Aei
|
i∈Aei| =
n k(n+1−k)
i∈A
ei∈Sn−1. Then0is the centroid of n∩u⊥A.
It suffices to show that 0 is the Santal ´o point ofPuA n. Let
el=PuAel, 1≤l≤n+1, E=[ei,i∈A], andF=[ej;j∈A].
ThenEandFare linear subspaces ofu⊥Asuch that dim(E)=k−1, dim(F)=n−k, x,y =0 for everyx∈Eandy∈F, and
SA:=conv ({ei,i∈A})⊂EandTA:=conv ({ej,j∈A})⊂F
are regular simplices with centre of mass at 0. It follows that 0 is the Santal ´o point ofSAinEand ofTAinF, when 0 is the Santal ´o point ofPuA n=conv(SA,TA). It
follows that 0 is the centroid of n∩u⊥A, which can be described as
n∩u⊥A=SA∗ ×TA∗,
whereS∗AandTA∗are the polars ofSAand ofTA, respectively, inEandF. A corollary of this result is the following.
PROPOSITION9.For every A⊂ {1, . . . ,n+1}, such that1≤k:=card(A)≤n, one has
||uA||CI( n,0)= ||uA||I( n,0).
Moreover, in the particular case when u⊥∩ n is itself a simplex, one has the following computational proposition.
PROPOSITION10.Let u∈Sn−1, and if u=n+1
i=1 uiei∈Sn−1 withn+1
i=1ui=0 and u1, . . . ,un≥0>un+1, then u⊥∩ nis a simplex and
ρI( n,0)(u)=vol( n∩u⊥)= 1 (n−1)!
(n+1)n+12 nn2−1
n 1
i=1(ui+n j=1uj) and
ρCI( n,0)(u)=vol(( n∩u⊥)∗0)∗s)= 1 (n−1)!
nn2+1 (n+1)n+21
n1
i=1ui.
It follows from Proposition 10 thatCI( n,0) has 2n+2 small faces aroundu= ±ei, 1≤i≤n+1. Nevertheless, it is easy to check that for such directionsu∈Sn−1 one has
1≤ vol( n∩u⊥) vol(( n∩u⊥)∗0)∗s) ≤ e
2.
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